Conditional density for $X$ given $X+Y=Z$ 
Suppose $X$ and $Y$ are i.i.d. random variables of exponential distribution with parameter $\lambda$. Let $Z=X+Y$. What is $f_X(x\mid Z=z)$?

By definition, the conditional density
$$
f_X(x\mid Z=z)=\frac{f_{X,Z}(x,z)}{f_Z(z)}.
$$
Using convolution, one can calculate $f_Z(z)$. How can I find the joint density $f_{X,Z}(x,z)$? I was trying
$$
f_{X,Z}(x,z)=f_Z(z\mid X=x)f_X(x).
$$
Intuitively, 
$$
f_Z(z\mid X=x)=f_Y(z-x).
$$
But I don't know how to justify it.
 A: Using CDF approach:
$$ \Pr\{Z \leq z|X = x\} = \Pr\{X + Y \leq z|X = x\} = \Pr\{Y \leq z - x\}$$
Then by differentiation with respect to $z$ you can immediately obtain the desired result.
A: You may usually treat probability density functions as analogous to probability mass functions.   (Taking care to avoid the Bertrand paradox and such.)
In this case, since $X,Y$ are iid and $Z=X+Y$ we have:


*

*$f_{X,Z}(x,z) = f_{X,Y}(x, z-x)$


[ $f_{X,Z}(x,z) = f_{Z\mid X}(z\mid x)f_X(x) = f_{Y\mid X}(z-x\mid x)f_X(x) = f_Y(z-x)f_X(x)$ ]


*$f_{Z}(z) = \int_\Bbb R f_{X,Y}(s, z-s)\operatorname d s$


Hence:
$$\begin{align}f_{X\mid Z}(x\mid z) & = \dfrac{f_{X,Z}(x,z)}{f_{Z}(z)} \;\mathbf 1_{z\in[0;\infty), x\in [0;z]}
\\[1ex] & = \dfrac{f_{X,Y}(x,z-x)}{\int_{\Bbb R} f_{X,Y}(s,z-s)\operatorname d s}\;\mathbf 1_{z\in[0;\infty), x\in [0;z]}
\\[1ex] & = \ldots
\end{align}$$
Which gives a rather nice result.   (Do you recognise the conditional density function of $X$ when $X+Y=z$?)
A: An exponential distribution describes the time between events in a Poisson process.
So, take a Poisson process with parameter $\lambda$, and start the clock.   Let $X$ be the time until the next Poisson event, and $X+Y$ be the time until the event following that.   $X,Y$ are thus independent and identically distributed exponential random variables.
Some time passes.   Just now the first event has occurred at time $x$ and we wish to know the probability that $Z$, the sum of times $X,Y$, is less than time $z$.   That is $\mathsf P(Z\leq z\mid X=x)$
We can't define it by Bayes' Rule, because $\mathsf P(X=x)=0$.   It's an almost impossible event.   However, it has just happened! 
Let us go back to basics.   The conditioned event will occur if and only if $Y\leq z-x$.
Thus the conditional probability that $X+Y\leq z$, when given that we know that $X$ is $x$ (and that $Y$ is not dependent on $X$), is just the probability that $Y\leq z-x$. 
Now we just have to determine the conditional probability density.
$$\mathsf P(Z\leq z\mid X=x) = \mathsf P(Y\leq z-x)
\\ \therefore f_{Z\mid X}(z\mid x) = f_Y(z-x)$$
A: Here is yet another approach for the justification you seek. Consider the following for arbitrary real numbers $a$ and $b$:
$$
\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}{\bf 1}_{z\leqslant b}f(z\mid x)\mathrm dz\right){\bf 1}_{x\leqslant a} f(x)\mathrm dx = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\bf 1}_{x\leqslant a,z\leqslant b}f(x,z)\mathrm dx\ \mathrm dz = {\bf P}[X\leqslant a, Z\leqslant b] = {\bf P}[X\leqslant a, X+Y\leqslant b] = {\bf P}[X\leqslant a, Y\leqslant b-X] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\bf 1}_{x\leqslant a,y\leqslant b-x}f(x,y)\mathrm dy\ \mathrm dx \\ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}{\bf 1}_{x\leqslant a,y\leqslant b-x}f(y)f(x)\mathrm dy\ \mathrm dx = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}{\bf 1}_{y\leqslant b-x}f(y)\mathrm dy\right)\ {\bf 1}_{x\leqslant a}f(x)\mathrm dx
$$
Most of the foregoing holds by definition; of course, above, use was also made of the independence of $X$ and $Y$. 
Now, both $\ \color{green}{\int_{-\infty}^{\infty}{\bf 1}_{z\leqslant b}f(z\mid x)\mathrm dz}\ $ and $\ \color{green}{\int_{-\infty}^{\infty}{\bf 1}_{y\leqslant b-x}f(y)\mathrm dy}\ $ are bounded, $\sigma(X)$-measurable functions.  So, given the arbitrariness of $a$, these functions are equal almost surely. That is,

$$
{\bf P}[Z\leqslant b\mid X=x]:=\int_{-\infty}^{\infty}{\bf 1}_{z\leqslant b}f(z\mid x)\mathrm dz \stackrel{a.s}{=} \int_{-\infty}^{\infty}{\bf 1}_{y\leqslant b-x}f(y)\mathrm dy = {\bf P}[Y\leqslant b-x]
$$

Following an analogous argument (which includes a change of variable) one can further show that
$$
f_Z(z\mid X=x)=\frac{\mathrm d}{\mathrm dz}{\bf P}[Z\leqslant z\mid X=x]\stackrel{a.s.}{=}\frac{\mathrm d}{\mathrm dz}{\bf P}[Y\leqslant z-x]=f_Y(z-x)
$$
